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26 Jul 2006, 08:03
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x^2 +2qx+ r =0 , ask for the numbers of intersection with X-axis.
1) q^2>r
2) r^2>q
According to the question bank the answer is A don't know how they arrived at the solution though
1) q^2>r
2) r^2>q
According to the question bank the answer is A don't know how they arrived at the solution though
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26 Jul 2006, 08:16
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The problem can restated as finding the points of intersection of y = x^2 +2qx+ r with x-axis. Note that on x-axis, y = 0 which means x^2 +2qx+ r = 0.
Note that the given equation is x^2 +2qx+ r =0 is a quadratic equation. "Quadratic" meaning the highest power of the variable x is 2. Hence, the equation has atmost 2 real solutions. The nature of the solutions is determined by the "discriminant" of the equation which is equal to b^2-4*a*c where the equation is ax^2 + b^x+c = 0.
The equation has
A. two distinct real roots if b^2-4ac > 0
B. has one real root if b^2-4ac = 0
C. has no real roots (imiginary roots) if b^2-4ac<0
In other words, the original problem is asking you to find out how many real roots does the given quadratic equation has (because if it has imiginary roots, then it will not intersect with X-axis).
So, by comparing the given equation x^2 +2qx+ r = 0 with the standard form ax^2 + b^x+c = 0, note that a =1 , b = 2q and c = r.
b^2-4ac = 4q^2-4r = 4(q^2-r).
So, the # of real roots of the given equation is determined by whether
q^2-r > 0 or =0 or less than zero.
Statement 1: q^2>r, hence, the given equation has two real roots. Which implies that it intersects with the X axis twice. Hence, statement 1 alone is sufficient.
Statement 2: r^2 > q implies r^4 > q^2, does not necessirily tell you anything about whether q^2 is greater than r. Hence, statement 2 alone is not sufficient.
Hence, the answer is A.
-mathguru
Note that the given equation is x^2 +2qx+ r =0 is a quadratic equation. "Quadratic" meaning the highest power of the variable x is 2. Hence, the equation has atmost 2 real solutions. The nature of the solutions is determined by the "discriminant" of the equation which is equal to b^2-4*a*c where the equation is ax^2 + b^x+c = 0.
The equation has
A. two distinct real roots if b^2-4ac > 0
B. has one real root if b^2-4ac = 0
C. has no real roots (imiginary roots) if b^2-4ac<0
In other words, the original problem is asking you to find out how many real roots does the given quadratic equation has (because if it has imiginary roots, then it will not intersect with X-axis).
So, by comparing the given equation x^2 +2qx+ r = 0 with the standard form ax^2 + b^x+c = 0, note that a =1 , b = 2q and c = r.
b^2-4ac = 4q^2-4r = 4(q^2-r).
So, the # of real roots of the given equation is determined by whether
q^2-r > 0 or =0 or less than zero.
Statement 1: q^2>r, hence, the given equation has two real roots. Which implies that it intersects with the X axis twice. Hence, statement 1 alone is sufficient.
Statement 2: r^2 > q implies r^4 > q^2, does not necessirily tell you anything about whether q^2 is greater than r. Hence, statement 2 alone is not sufficient.
Hence, the answer is A.
-mathguru
